$$$t \sin{\left(t \right)} \cos{\left(t \right)}$$$ 的积分

求$$$\int t \sin{\left(t \right)} \cos{\left(t \right)}\, dt$$$。

对于积分$$$\int{t \sin{\left(t \right)} \cos{\left(t \right)} d t}$$$，使用分部积分法$$$\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$$$。

设 $$$\operatorname{u}=t$$$ 和 $$$\operatorname{dv}=\sin{\left(t \right)} \cos{\left(t \right)} dt$$$。

则 $$$\operatorname{du}=\left(t\right)^{\prime }dt=1 dt$$$ (步骤见 »)，并且 $$$\operatorname{v}=\int{\sin{\left(t \right)} \cos{\left(t \right)} d t}=\frac{\sin^{2}{\left(t \right)}}{2}$$$ (步骤见 »)。

因此，

$${\color{red}{\int{t \sin{\left(t \right)} \cos{\left(t \right)} d t}}}={\color{red}{\left(t \cdot \frac{\sin^{2}{\left(t \right)}}{2}-\int{\frac{\sin^{2}{\left(t \right)}}{2} \cdot 1 d t}\right)}}={\color{red}{\left(\frac{t \sin^{2}{\left(t \right)}}{2} - \int{\frac{\sin^{2}{\left(t \right)}}{2} d t}\right)}}$$

对 $$$c=\frac{1}{2}$$$ 和 $$$f{\left(t \right)} = \sin^{2}{\left(t \right)}$$$ 应用常数倍法则 $$$\int c f{\left(t \right)}\, dt = c \int f{\left(t \right)}\, dt$$$：

$$\frac{t \sin^{2}{\left(t \right)}}{2} - {\color{red}{\int{\frac{\sin^{2}{\left(t \right)}}{2} d t}}} = \frac{t \sin^{2}{\left(t \right)}}{2} - {\color{red}{\left(\frac{\int{\sin^{2}{\left(t \right)} d t}}{2}\right)}}$$

应用降幂公式 $$$\sin^{2}{\left(\alpha \right)} = \frac{1}{2} - \frac{\cos{\left(2 \alpha \right)}}{2}$$$，并令 $$$\alpha=t$$$:

$$\frac{t \sin^{2}{\left(t \right)}}{2} - \frac{{\color{red}{\int{\sin^{2}{\left(t \right)} d t}}}}{2} = \frac{t \sin^{2}{\left(t \right)}}{2} - \frac{{\color{red}{\int{\left(\frac{1}{2} - \frac{\cos{\left(2 t \right)}}{2}\right)d t}}}}{2}$$

对 $$$c=\frac{1}{2}$$$ 和 $$$f{\left(t \right)} = 1 - \cos{\left(2 t \right)}$$$ 应用常数倍法则 $$$\int c f{\left(t \right)}\, dt = c \int f{\left(t \right)}\, dt$$$：

$$\frac{t \sin^{2}{\left(t \right)}}{2} - \frac{{\color{red}{\int{\left(\frac{1}{2} - \frac{\cos{\left(2 t \right)}}{2}\right)d t}}}}{2} = \frac{t \sin^{2}{\left(t \right)}}{2} - \frac{{\color{red}{\left(\frac{\int{\left(1 - \cos{\left(2 t \right)}\right)d t}}{2}\right)}}}{2}$$

逐项积分：

$$\frac{t \sin^{2}{\left(t \right)}}{2} - \frac{{\color{red}{\int{\left(1 - \cos{\left(2 t \right)}\right)d t}}}}{4} = \frac{t \sin^{2}{\left(t \right)}}{2} - \frac{{\color{red}{\left(\int{1 d t} - \int{\cos{\left(2 t \right)} d t}\right)}}}{4}$$

应用常数法则 $$$\int c\, dt = c t$$$，使用 $$$c=1$$$：

$$\frac{t \sin^{2}{\left(t \right)}}{2} + \frac{\int{\cos{\left(2 t \right)} d t}}{4} - \frac{{\color{red}{\int{1 d t}}}}{4} = \frac{t \sin^{2}{\left(t \right)}}{2} + \frac{\int{\cos{\left(2 t \right)} d t}}{4} - \frac{{\color{red}{t}}}{4}$$

设$$$u=2 t$$$。

则$$$du=\left(2 t\right)^{\prime }dt = 2 dt$$$ (步骤见»)，并有$$$dt = \frac{du}{2}$$$。

因此，

$$\frac{t \sin^{2}{\left(t \right)}}{2} - \frac{t}{4} + \frac{{\color{red}{\int{\cos{\left(2 t \right)} d t}}}}{4} = \frac{t \sin^{2}{\left(t \right)}}{2} - \frac{t}{4} + \frac{{\color{red}{\int{\frac{\cos{\left(u \right)}}{2} d u}}}}{4}$$

对 $$$c=\frac{1}{2}$$$ 和 $$$f{\left(u \right)} = \cos{\left(u \right)}$$$ 应用常数倍法则 $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$：

$$\frac{t \sin^{2}{\left(t \right)}}{2} - \frac{t}{4} + \frac{{\color{red}{\int{\frac{\cos{\left(u \right)}}{2} d u}}}}{4} = \frac{t \sin^{2}{\left(t \right)}}{2} - \frac{t}{4} + \frac{{\color{red}{\left(\frac{\int{\cos{\left(u \right)} d u}}{2}\right)}}}{4}$$

余弦函数的积分为 $$$\int{\cos{\left(u \right)} d u} = \sin{\left(u \right)}$$$：

$$\frac{t \sin^{2}{\left(t \right)}}{2} - \frac{t}{4} + \frac{{\color{red}{\int{\cos{\left(u \right)} d u}}}}{8} = \frac{t \sin^{2}{\left(t \right)}}{2} - \frac{t}{4} + \frac{{\color{red}{\sin{\left(u \right)}}}}{8}$$

回忆一下 $$$u=2 t$$$:

$$\frac{t \sin^{2}{\left(t \right)}}{2} - \frac{t}{4} + \frac{\sin{\left({\color{red}{u}} \right)}}{8} = \frac{t \sin^{2}{\left(t \right)}}{2} - \frac{t}{4} + \frac{\sin{\left({\color{red}{\left(2 t\right)}} \right)}}{8}$$

因此，

$$\int{t \sin{\left(t \right)} \cos{\left(t \right)} d t} = \frac{t \sin^{2}{\left(t \right)}}{2} - \frac{t}{4} + \frac{\sin{\left(2 t \right)}}{8}$$

加上积分常数：

$$\int{t \sin{\left(t \right)} \cos{\left(t \right)} d t} = \frac{t \sin^{2}{\left(t \right)}}{2} - \frac{t}{4} + \frac{\sin{\left(2 t \right)}}{8}+C$$

$$$\int t \sin{\left(t \right)} \cos{\left(t \right)}\, dt = \left(\frac{t \sin^{2}{\left(t \right)}}{2} - \frac{t}{4} + \frac{\sin{\left(2 t \right)}}{8}\right) + C$$$A